int frac{dx}{cos x(1+cos x)} = | vedclass

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Question

(C) Let $I = \int \frac{dx}{\cos x(1+\cos x)}$.We can write the integrand as $\frac{1}{\cos x(1+\cos x)} = \frac{1+\cos x - \cos x}{\cos x(1+\cos x)}

Vedclass · Accepted Answer

$\log |\sec x + 	an x| - 	an \left(\frac{x}{2}ight) + c$,where $c$ is the constant of integration

Vedclass · Answer

(C) Let $I = \int \frac{dx}{\cos x(1+\cos x)}$.We can write the integrand as $\frac{1}{\cos x(1+\cos x)} = \frac{1+\cos x - \cos x}{\cos x(1+\cos x)} = \frac{1}{\cos x} - \frac{1}{1+\cos x}$.Now,$I = \int \sec x \, dx - \int \frac{dx}{1+\cos x}$.We know that $\int \sec x \, dx = \log |\sec x + 	an x| + c_1$.For the second part,use the identity $1+\cos x = 2 \cos^2 \left(\frac{x}{2}ight)$.So,$\int \frac{dx}{1+\cos x} = \int \frac{dx}{2 \cos^2 \left(\frac{x}{2}ight)} = \frac{1}{2} \int \sec^2 \left(\frac{x}{2}ight) dx$.Integrating this,we get $\frac{1}{2} \cdot \frac{	an(x/2)}{1/2} = 	an \left(\frac{x}{2}ight) + c_2$.Combining these,$I = \log |\sec x + 	an x| - 	an \left(\frac{x}{2}ight) + c$.

$\int \frac{dx}{\cos x(1+\cos x)} = $

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